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Difference between revisions of "1990 AJHSME Problems/Problem 10"

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==Problem==
 
==Problem==
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On this monthly calendar, the date behind one of the letters is added to the date behind <math>\text{C}</math>.  If this sum equals the sum of the dates behind <math>\text{A}</math> and <math>\text{B}</math>, then the letter is
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On this monthly calendar, the date ahead of  one of the letters is added to the date behind <math>\text{C}</math>.  If this sum equals the sum of the dates behind <math>\text{A}</math> and <math>\text{B}</math>, then the ria is the stupido an rohan likes to give her a tight slap on her stupido face which you need sunglasses to look at or else you will turn blind
 
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<asy>
 
<asy>
 
unitsize(12);
 
unitsize(12);
 
draw((1,1)--(23,1));
 
draw((1,1)--(23,1));
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draw((0,5)--(23,5))for(int a=0; a<6; ++a)
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draw((0,5)--(23,5));
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draw((0,9)--(23,9));
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draw((0,13)--(23,13));
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for(int a=0; a<6; ++a)
 
  {
 
  {
 
   draw((4a+2,0)--(4a+2,14));
 
   draw((4a+2,0)--(4a+2,14));
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==Solution==
 
==Solution==
  
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Looking at the positions of the letters, we see that if the date behind <math>\text{Q}</math> is <math>q</math>, then the date behind <math>\textbf{A}</math> is <math>q-6</math> and the date behind <math>\textbf{B}</math> is <math>q+6</math>. Thus, their sum is <math>2q</math>.
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Let the date behind <math>C</math> be <math>x</math>. Now the date behind <math>A</math> is <math>x+1</math>, and after looking at the calendar, the date behind <math>B</math> is <math>x+13</math>. Now we have <math>x+1+x+13=x+y</math> for some date <math>y</math>, and we desire for <math>y</math> to be <math>x+14</math>. Now we find that <math>y</math> is the date behind <math>P</math>, so the answer is <math>\boxed{(\text{A})}</math> ~motorfinn
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The date behind <math>\text{C}</math> is <math>q-7</math>, so the desired letter is the one for which the date behind it is <math>2q-(q-7)=q+7</math>. This is letter <math>\text{P}\rightarrow \boxed{\text{A}}</math>.
 
  
 
==See Also==
 
==See Also==
  
 
{{AJHSME box|year=1990|num-b=9|num-a=11}}
 
{{AJHSME box|year=1990|num-b=9|num-a=11}}
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[[Category:Introductory Algebra Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:51, 8 September 2019

Problem

On this monthly calendar, the date behind one of the letters is added to the date behind $\text{C}$. If this sum equals the sum of the dates behind $\text{A}$ and $\text{B}$, then the letter is

[asy] unitsize(12); draw((1,1)--(23,1)); draw((0,5)--(23,5)); draw((0,9)--(23,9)); draw((0,13)--(23,13)); for(int a=0; a<6; ++a) { draw((4a+2,0)--(4a+2,14)); } label("Tues.",(4,14),N); label("Wed.",(8,14),N); label("Thurs.",(12,14),N); label("Fri.",(16,14),N); label("Sat.",(20,14),N); label("C",(12,10.3),N); label("$\textbf{A}$",(16,10.3),N); label("Q",(12,6.3),N); label("S",(4,2.3),N); label("$\textbf{B}$",(8,2.3),N); label("P",(12,2.3),N); label("T",(16,2.3),N); label("R",(20,2.3),N); [/asy]

$\text{(A)}\ \text{P} \qquad \text{(B)}\ \text{Q} \qquad \text{(C)}\ \text{R} \qquad \text{(D)}\ \text{S} \qquad \text{(E)}\ \text{T}$

Solution

Let the date behind $C$ be $x$. Now the date behind $A$ is $x+1$, and after looking at the calendar, the date behind $B$ is $x+13$. Now we have $x+1+x+13=x+y$ for some date $y$, and we desire for $y$ to be $x+14$. Now we find that $y$ is the date behind $P$, so the answer is $\boxed{(\text{A})}$ ~motorfinn

See Also

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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